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Geometric series algebra / exponential/ 2 summations

  1. Nov 5, 2016 #1
    The problem statement, all variables and given/known data

    I want to show that ## \sum\limits_{n=1}^{\infty} log (1-q^n) = -\sum\limits_{n=1}^{\infty}\sum\limits_{m=1}^{\infty} \frac{q^{n.m}}{m} ##, where ##q^{n}=e^{2\pi i n t} ## , ##t## [1] a complex number in the upper plane.


    2. Relevant equations

    Only that ## e^{x} = \sum\limits_{m=0}^{\infty} \frac{x^{m}}{m!}## [2]

    3. The attempt at a solution

    I can see that both series start from ##n,m=1## in the RHS of [1] , so if I use [2] starting from ##m=1## will result in the '1' cancelling as needed i.e. I have ## \sum\limits_{n=1}^{\infty} log (1-q^n) = \sum\limits_{n=1}^{\infty} log (\sum\limits_{m=1}^{\infty} \frac{(2\pi int)^{m}}{m!})##

    I don't really no where to go now, I don't see how you can get another expansion in terms of ##q^{m}## from this to give the required ##q^{n.m}##, I can see there's a minus sign too but I'm struggling to use this as a clue as well.

    Many thanks in advance.
     
  2. jcsd
  3. Nov 5, 2016 #2
    I am also struggling to show that ## \sum\limits_{n=1}^{\infty} n \frac{q^{n}}{(1-q^n)} = \sum\limits_{n=1}^{\infty} n \sum\limits_{m=1}^{\infty} q^{m.n} ##

    I'm guessing it's the same things I'm missing in both cases...many thanks in advance.
     
  4. Nov 5, 2016 #3
    Unless I'm missing something, isn't this just a matter of Taylor expanding ##\log(1-q^{n})##?
     
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